If $P(A\cup B)=P(A\cap B)$ then find the relation between $P(A)$ and $P(B)$.
A. $P(A)=P(\bar{A})$
B. $P(A\cap B)=P(A'\cap B')$
C. $P(A)=P(B)$
D. None of these

Hint: Apply the formula $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ and check each and every option by substituting it in the formula.

Formula used: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Complete step by step solution: We know that $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ ………. (1)
If $P(A)=P(\bar{A})$then putting this in equation (1) we get,
$\Rightarrow P(A'\cup B)=P(A')+P(B)-P(A'\cap B)$
$\Rightarrow P(A\cup B)\ne P(A\cap B)$
Therefore, option A. is not true.

If $P(A\cap B)=P(A'\cap B')$ then putting this in equation (1) we get,
$\Rightarrow P(A\cup B)=P(A)+P(B)-P(A'\cap B')$
$\Rightarrow P(A\cup B)\ne P(A\cap B)$
Therefore, option B. is also not true.

If $P(A)=P(B)$ then putting this in equation (1) we get,
$\begin{align}
  & \Rightarrow P(A)=2P(A)-P(A) \\
 & \Rightarrow P(A)=P(A) \\
 & \Rightarrow P(A\cup B)=P(A\cap B) \\
\end{align}$

Thus, Option (C) is correct.

Note: Use can also use the formula $P(A\cap B)=P(A)+P(B)-P(A\cup B)$ in a similar way as shown in the solution as you will get the same results. And don’t get confused with $\bar{A}$and $ A'$ as they both have the same meaning. Don’t make silly mistakes while solving equations.